Evaluate the following integral:

assume $\mathrm{x}=\frac{1}{\mathrm{t}}$

$d x=-\frac{1}{t^{2}} d t$

$-\int \frac{t d t}{\left(t^{2}+1\right)\left(\sqrt{t^{2}-1}\right.}$

Let $\mathrm{t}^{2}-1=\mathrm{u}^{2}$'

$\mathrm{tdt}=\mathrm{udu}$

$-\int \frac{u d u}{\left(u^{2}+2\right) u}$

$-\int \frac{d u}{\left(u^{2}+2\right)}$

Using identity $\int \frac{1}{x^{2}+1} d x=\arctan (x)$

$-\frac{1}{\sqrt{2}} \arctan \left(\frac{\mathrm{u}}{\sqrt{2}}\right)+\mathrm{c}$

Substituting $u=\sqrt{t^{2}-1}$

$-\frac{1}{\sqrt{2}} \arctan \left(\frac{\sqrt{t^{2}-1}}{\sqrt{2}}\right)+c$

Substituting $\mathrm{t}=\frac{1}{\mathrm{x}}$

$-\frac{1}{\sqrt{2}} \arctan \left(\frac{\sqrt{\frac{1}{x^{2}}-1}}{\sqrt{2}}\right)+c$